Here on earth and also on the moon two identical heavy crates (300kg) sit at rest on 2 identical level platforms with coefficients of static friction equal to (0.90). If one end of both platforms is slowly raised at the same time in both locations, will the crates eventually break free of the static friction at the same time or, because of the fact that gravity differs significantly between the two planets, will one break free sooner (i.e. at a lower angle of incline) than the other? If so, on which planet will the crate break free sooner?
The angle A at which a crate on a tilted platform "breaks free" is given by
M g *mus* cos A = M g sin A
The right side is the actual gravity force component down the ramp and the left side is the maximum static friction force to resist motion. The "g" is the local value, which is 1/6 as large on the Moon.
The solution is
tan A = mus (the static friction coefficient).
Therefore the angle A will be the same for Earth or moon. Both M and the "local" g cancel
To determine whether the crates will break free of the static friction at the same time or at different angles of incline on Earth and the Moon, we need to consider the effects of gravity on the frictional force.
First, let's calculate the maximum static friction force that can act on the crates. The maximum static friction force can be determined using the equation:
F_static_max = µ_s * N
where µ_s is the coefficient of static friction and N is the normal force acting on the crates.
On both Earth and the Moon, the normal force is equal to the weight of the crates.
Weight = m * g
where m is the mass of the crates and g is the acceleration due to gravity.
Since the mass of the crates is the same on Earth and the Moon (300 kg), the weight of the crates will differ due to the different values of acceleration due to gravity.
Now, let's calculate the weight and the maximum static friction force on both Earth and the Moon.
On Earth:
Weight = 300 kg * 9.8 m/s^2 = 2940 N
F_static_max = 0.9 * 2940 N = 2646 N
On the Moon:
Weight = 300 kg * 1.6 m/s^2 = 480 N
F_static_max = 0.9 * 480 N = 432 N
As we can see, the maximum static friction forces differ between Earth (2646 N) and the Moon (432 N) due to the difference in gravity.
Now, let's consider the scenario of raising one end of both platforms slowly. As the angle of incline increases, the component of the weight acting parallel to the plane of the platform (mg sinθ) also increases.
The force parallel to the plane is given by: F_parallel = mg * sinθ
Once the force parallel to the plane exceeds the maximum static friction force, the crates will break free. In this case, if the force parallel to the plane is equal to or greater than the maximum static friction force, the crates will start to move.
Since the force parallel to the plane (mg sinθ) will be the same for both Earth and the Moon, the crates will break free at the same angle of incline on both planets.
Therefore, the crates will break free of the static friction at the same time and at the same angle of incline on Earth and the Moon.